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In a certain Algebra 2 class of 29 students, 18 of them play basketball and 6 of them play baseball. There are 7 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball? Answer:

In a certain Algebra 22 class of 2929 students, 1818 of them play basketball and 66 of them play baseball. There are 77 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?\newlineAnswer:

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Q. In a certain Algebra 22 class of 2929 students, 1818 of them play basketball and 66 of them play baseball. There are 77 students who play neither sport. What is the probability that a student chosen randomly from the class plays both basketball and baseball?\newlineAnswer:
  1. Calculate Total Students Playing Sports: Let's first determine the total number of students who play at least one sport. Since there are 77 students who play neither sport, the number of students who play at least one sport is the total number of students minus those who play neither.\newlineCalculation: 2929 students (total) - 77 students (play neither) = 2222 students (play at least one sport).
  2. Apply Inclusion-Exclusion Principle: Now, let's use the principle of inclusion-exclusion to find out how many students play both sports. The principle states that the number of students playing either basketball or baseball is equal to the sum of students playing each sport minus the number of students playing both sports.\newlineCalculation: Let BB be the number of students playing basketball, let BaBa be the number of students playing baseball, and let B∩BaB \cap Ba be the number of students playing both. We have B=18B = 18, Ba=6Ba = 6, and we need to find B∩BaB \cap Ba.\newlineAccording to the principle of inclusion-exclusion:\newlineB+Ba−B∩Ba=B + Ba - B \cap Ba = Number of students playing at least one sport.\newline18+6−B∩Ba=2218 + 6 - B \cap Ba = 22.
  3. Find Students Playing Both Sports: Solving the equation from the previous step for B∩BaB\cap Ba gives us the number of students playing both basketball and baseball.\newlineCalculation: 18+6−B∩Ba=2218 + 6 - B\cap Ba = 22\newline24−B∩Ba=2224 - B\cap Ba = 22\newlineB∩Ba=24−22B\cap Ba = 24 - 22\newlineB∩Ba=2B\cap Ba = 2.
  4. Calculate Probability: Finally, we calculate the probability that a student chosen randomly from the class plays both basketball and baseball. The probability is the number of students playing both sports divided by the total number of students.\newlineCalculation: Probability = B∩Ba/Total number of studentsB\cap Ba / \text{Total number of students}\newlineProbability = 229\frac{2}{29}.

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